To a (projective smooth) algebraic surface $S$ over an algebraically closed field and a divisor $D$ of $S$, we can associate $n= \dim |D|$ and $g$, the genus of a generic member of $|D|$. I would like to fix $g$ and vary $S,D$ so as to make $n$ as large as possible. Is $n$ unbounded and, if not, what is the optimal bound? I am mainly interested in the case of positive characteristic but an answer over the complex numbers would be welcome too.

2 Answers
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EDIT: my example did not work, as pointed out by quim in the comments.
The example he suggests however works: take $S$ the blowup of $P^2$ at a point $x$ and $D$ the strict transform of a curve of degree $d$ with a singular point of multiplicity $d-1$ at $x$. The general $D$ is smooth and $|D|$ has dimension $2d$.\
Of course, a similar construction can be used to construct examples with $g>0$.

On the other hand, if the Kodaira dimension of $S$ is $\ge 0$ and $D$ is irreducible, then $n=\dim|D|$ is bounded by $g$ by the following argument.

Up to blowing up $S$ we may assume that the general $D$ is smooth.
Let $m>0$ be such that $mK_S\ge 0$. If $n>0$, then $D$ is not in the fixed part of $|mK_S|$, hence $K_SD\ge 0$. Hence by the adjunction formula $D|_D$ is a divisor of $D$ of degree $\le 2g-2$ and therefore it satisfies $2\dim|D|_D|\le \deg D$ (this is Clifford's theorem if $D$ is special and it is trivially true otherwise).
So we have $\dim |D|\le \dim|D|_D|+1\le D^2/2+1\le (D^2+K_SD)/2+1=g$.

I, hence $K_S|_D$ is effective and $D|_D$ is special. Then Clifford's theorem and the adjunction formula give $2(n-1)\le D^2\le D^2+K_SD=2g-2$.

I don't understand the plane example. The strict transform of C on S has selfintersection $d^2-\frac{(d-1)(d-2)}{2}4<0$ so the dimension of |D| is zero. I guess what grows as d^3/6 must be the dimension of the (nonlinear) family of rational curves of degree d?
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quimNov 11 '10 at 22:07

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On the other hand, you can construct a similar example with the linear family of plane curves of degree d with a point of multiplicity d-1. On the blowup, dim |D|=2d and the genus is 0.
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quimNov 11 '10 at 22:15

I'm sorry, I made a mistake in computing the dimension of $|D|$ (I was thinking of $P^3$ instead of $P^2$), but your example works. The second part of my answer is ok, I think.
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ritaNov 11 '10 at 22:26

The following example shows that $n$ can be unbounded, for any value of $g$, even if $S$ is fixed.

Let $C$ be a smooth curve of genus $g$, and let $\mathcal{E}$ be a normalized rank $2$ vector bundle on $E$. Let $\mathfrak{e}$ be the divisor on $C$ such that $\bigwedge^2 \mathcal{E}=\mathcal{O}_C(\mathfrak{e})$, and consider the projective bundle

$S:=\mathbb{P}(\mathcal{E}) \stackrel{\pi}{\longrightarrow} C$.

Let $C_0$ be a section such that $C_0^2 = \deg \mathfrak{e}$, and let $f$ be the class of a fiber of $\pi$.

Assume now that $\mathfrak{b}$ is any divisor on $C$ having the following properties:

$\mathfrak{b}$ is nonspecial;

$|\mathfrak{b}|$ and $|\mathfrak{b}+ \mathfrak{e}|$ have no base points.

Notice that these conditions are satisfied as soon as $\mathfrak{b}$ is a general divisor of sufficiently high degree.

By [Hartshorne, Algebraic geometry, Ex. 2.11 p. 385], there exists a section
$D$ linearly equivalent to $C_0 + \mathfrak{b} f$, and moreover $|D|$ is base-point free. By Bertini's theorem it follows that the general element of $|D|$ is a smooth curve of genus $g$, and the dimension of $|D|$ clearly goes to infinity when $\deg \mathfrak{b}$ goes to infinity.

A simplified version of your example consists in taking $S=P^1\times C$ and $D=f_1+df_2$, where $f_i$ is the fiber of the projection of $S$ on the $i$-th factor. If $d$ is large enough, then $|D|$ is free of dimension $n=2d+1-g$ and the general $D$ is smooth of genus $g$.
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ritaNov 12 '10 at 13:10